A tangent galvanometer has a coil of 50 turns and a radius of 20cm. The horizontal component of the earth's magnetic field is `B_H = 3 xx 10^(-5) T`. Find the current which gives a diflection of `45^@)`.

Explanation

We have
`I = K tantheta = (2rB_H)/(mu_0 n) tantheta`
`= (2 xx (0.20 m) xx ( 3 xx 10^(-5) T)/(4pi xx 10^(-7) T m A^(-1) xx 50) tan 45^@ = 0.19 A`.